POLAR CO-ORDINATES
We often use the Cartesian Coordinate system to define the position of a point on a plane.
However, it’s not the only way. In this section, we would introduce another system, the polar
coordinate.
1. Representation
Just like the Cartesian coordinate system, we use two real numbers (r,  ) to represent the
position of a point on the plane in the polar coordinate system. Here, the magnitude of r is the
distance of the point from the origin and  is its angular position, measured anti-clockwisely,
relative to an axis, usually called the initial line. If r is positive, the segment joining (r,  ) and the
origin make an angle  in the anti-clockwise direction with the initial line. Conversely, if r is
negative, it is in the other direction of the segment relative to the origin.
Illustration. The point (1,
3
 ) in polar coordinate system is shown in figure 1. The line with an
4
arrow is the initial line.
3
(1,  )
4
1
3

4
3
 
4
2
3
1
(2,   ) or ( 2,  )
4
4
Figure 1: Examples of polar coordinate
Page 1 of 11
Note that in the polar coordinate system, the representation of a point is not unique. For
instance, the (r,  ) and (r,   2 ) are the same point for any r and  , and (0,  ) is always the
origin.
Usually, we will take the positive x-axis as the initial line, with the origin coincident with the
one in Cartesian Plane and the distance meaning the same as that in the Cartesian coordinate system.
It can be seen that a point with polar coordinate (r,  ) has Cartesian coordinate ( r cos  , r sin  ).
The corresponding polar coordinate of a point (x, y) in the Cartesian coordinate representation is
(r,  ), where
 / 2
3 / 2

r  x 2  y 2 ,  
arctan( y / x)
  arctan( y / x)
, y  0, x  0
, y  0, x  0
,x 0
,x 0
Here, arctan maps any real number to its corresponding principal value , i.e. arctan :
Illustration. The points with polar coordinates (2,
 (
 
, )
2 2

3
), ( 2 ,
) and (3,  ) has Cartesian
2
6
coordinate ( 3, 1) , (1,1) and (3,0) respectively.
2. Graphs in Polar Coordinates
Like Cartesian coordinate system, we can describe the solution of an equation f (r , )  0 by
a graph in the polar coordinate system. Two of the simplest types should be circles centred at the
origin and straight lines passing through the origin. They have equations
r  a and   
respectively. Here a  0 is the radius of the circle, and  is the inclination of the line.
Example 2.1.
Sketch the graph of the following equation:
a. r   , 0    2
b. r  2 cos  , 0    
c. r  1  cos  ,  
Solution.
The graph (a) and (c) may be new to you. When expressed in Cartesian coordinate, their equations
Page 2 of 11
are quite complicated. Such type of curves is called spiral of Archimedes and cardioids
respectively. Graph (b) is of course a circle. The sketches of them are as follow:
1
1
0.5
-2
2
4
6
-1
0.5
1
1.5
2
-2
-0.5
-3
-4
-1
figure 2:
r 
figure 3:
r  2cos
1
0.5
-2
-1.5
-1
-0.5
-0.5
-1
figure 4:
r  1  cos
Example 2.2.
Sketch the graph of the following polar equations:
a. r  1  2sin  , 0    2
b. r  4sin 3 , 0    
Solution.
2
-1.5
-1
-0.5
0.5
1
1.5
1
-0.5
-1
figure 5:
-3
-2
-1
1
-1.5
-1
-2
-2
-2.5
-3
-3
-4
r  1  2sin 
figure 6:
2
3
r  4sin3
Page 3 of 11
The graph of (a) is called a limacon while that of (b) is called a 3-leafed rose. Generally n-leafed
roses are named in a way similar as suggested by their shapes.
It is always easy to transform a equation in Cartesian coordinate to the polar one by directly
substituting x  r cos , y  r sin  . For example, the straight line 2 x  y  3 can be transformed
3
to 2r cos  3r sin   3 , or r 
for all  
such that the denominator not
2 cos   3sin 
equal to 0. However, the reverse process may be complicated, as can be seen from the complexity
of the function that transforms (r ,  ) to the corresponding ( x, y ) given before.
When we want to find the intersections of two curves, we cannot do it merely by solving the
corresponding system of equations. In this way, we may only find some of the intersection points. It
will be illustrated in the example 2.3.
Example 2.3.
Let C1 : r  2  5cos  and C2 : r  2  2sin  , where 0    2 .
a.
Solve the system of equations of the two curves.
b.
Sketch the graphs of the curves.
c.
Find all the intersection point of the curves.
Solution.
5
5
From the two given equations, 2  5cos  r  2  2sin  or tan    . i.e.     arctan( )
2
2
5
or   2  arctan( ) . Substitute them back to the original equation, getting the solutions
2
10
5
10
5
(2 
,   arctan( ))  (0.14, 0.62 ) and (2 
, 2  arctan( ))  (3.86,1.62 ) .
2
2
29
29
4
2
-2
2
4
6
-2
-4
figure 7: C1 : r  2  5 cos  and C2 : r  2  2 sin 
Page 4 of 11
From the graphs, there should be 5 intersection points. From the equations, it is clear that both
curves pass through the origin. The remaining two intersections correspond to the root of the
equation [2  5cos(   )]  2  2sin  or 5cos  sin   4 . Solving it we find that the two
remaining points are
1
1


266  40 10 , cos 1[ (20  10)]   (0.48, 0.28 ) and
2
26
 13

1
1


266  40 10 , 2  cos 1[ (20  10)]   (2.91,1.85 ) .
2
13
26


Example 2.3 show that we may not be able to find all the intersection points simply by solving
the equations. The reason is that the representation of a point is not unique. For example, the two
2
2
curves C1 in the example passes through the origin at (0, cos 1 (  )) and (0,   cos 1 (  )) ,
5
5

while C2 passes it at (0, ) only. Hence, the intersection point can’t be found by solving the
2
equations. Similarly, although we have restricted the domain in [0, 2 ) , points other than the origin
also have two representations, namely (r ,  ) and (r ,    ) . This explain why we use the
equation [2  5cos(   )]  2  2sin  to find the remaining points.
Exercise

1. Sketch the graph of the equations r  3  2cos and r  3  3sin(  ) .
4
2. The polar equation r  a  b cos with (a, b)  (0, 0) may represent a circle, a limacon or a
cardioid. State the necessary and sufficient conditions for a and b such that they occur.
3. Measurements of Curves in Polar Equation
Theorem 3.1.
Let r  r ( ) be the polar equation of a curve. Then the slope of the tangent at (r ,  ) is given by:
dr
tan 
r
dy
d

dx dr  r tan 
d
Page 5 of 11
Proof. x  r cos , y  r sin  , so
dx
dr
dy
dr
 cos 
 r sin  and
 sin 
 r cos  . Hence,
d
d
d
d
dy
dr
dr
sin 
 r cos  tan 
r
dy d
d
d
.



dx dx cos  dr  r sin  dr  r tan 
d
d
d
Q.E.D.
Theorem 3.2.
Let r  r ( ) be the polar equation of a curve. Let S be the region consisting of points (r ,  )
satisfying 0  r  r ( ) or r ( )  r  0 , and 1     2 . Then the area of S is given by:
Area ( S ) 
1 2
r ( ) 2 d


2 1
Example 3.1.
Find the area of the region bounded by the circle r  2a cos  , a  0 .
Solution.
Here a is the radius of the circle centred at (a, 0) . The region enclosed by the circle is
S  {(r ,  ) | 0  r  2a cos  , 0     }

1 
1
(2a cos  ) 2 d 2a 2  cos 2  d  2a 2 (  )   a 2 .

0
2 0
2
Note that if we use [0, 2 ] to be the range of  , almost every point in S is “considered” twice, so
Hence, by theorem 3.2, the area is
yielding the wrong answer 2 a 2 instead.
Example 3.2.
Find the area enclosed between the two loops of the limacon 1  2cos , where  
.
Solution.
The idea is first to find out an interval of the domain corresponding to each loop and then we can
apply theorem 3.2 to calculate the area of them. The difference is the answer we want.
From figure 8, the large loop correspond to r  0 and the small loop correspond to r  0 , which
2
4
2 4
means   [0, ]  [ , 2 ] and   [ , ] respectively.
3
3
3 3
Page 6 of 11
1.5
1
0.5
0.5
1
1.5
2
2.5
3
-0.5
-1
-1.5
figure 8: r  1  2 cos 
Thus, their areas are
1 2 3
1 2
(1  2 cos  ) 2 d   (1  2 cos  ) 2 d

2 0
2 4 3

2 3

2 3

2 3
0
0
0
(1  2 cos  ) 2 d
(1  4 cos   4 cos 2  )d
(1  4 cos   2  2 cos 2 )d
 3  4sin   sin 2
 2 
2 3
0
3 3
2
and
1 4 3
(1  2 cos  ) 2 d

2

3
2
4 3
1
 (3  4sin   sin 2 )
2
2 3
 
3 3
2
accordingly. Hence, the required area is (2 
3 3
3 3
)  ( 
)  3 3.
2
2
Next, we will discuss the arc length of a curve.
Page 7 of 11
Theorem 3.3.
Let r  r ( ) be the polar equation of a curve, a  b be real numbers. The arc length s of the
curves form   1 to    2 is given by:
s
2
1
2
 dr 
2

  r d
d



Example 3.3.
Find the perimeter of the circle r  2a cos  , a  0 in example 3.1.
Solution.
Notice that  ranges from 0 to  , r ( ) runs round the circle once, and if we take 1  0 ,
2   , then the curve r ( ) would repeat a ‘path’ of positive length. Therefore, we should take
1  0, 2   in our calculation. The arc length is


0
[

d
(2a cos  )]2  (2a cos  ) 2 d   (2a sin  ) 2  (2a cos  ) 2 d
0
d

  2ad
0
 2 a
Example 3.4.
Find the arc length of the 5-leafed rose with polar equation r  sin5 .
Solution.
Just like in example 3.3, we can check that we should take 1  0 , 2   to produce to correct
answer. Thus, the arc length of the curves is


0
[

d
(sin 5 )]2  (sin 5 ) 2 d   25cos 2 (5 )  sin 2 (5 ) d
0
d




0
0
1  24 cos 2 (5 )d
1  24 cos 2 (5 )d
 10.51
The integral in the second last line is known as elliptic integral, which cannot be computed easily.
The approximation 10.51 is found by some numerical method instead.
Page 8 of 11
Example 3.5.
Let C1 : r  1  sin  , C2 : r  1  cos  be curves expressed in polar coordinate, where  ranges
over all real numbers. Find the arc length of the part of C2 which lies in the region bounded by
C1 .
Solution.
2
1.5
1
0.5
-2
-1.5
-1
-0.5
0.5
1
-0.5
-1
figure 9: C1 : r  1  sin  and C2 : r  1  cos 
Both C1 and C2 are cardioids. First of all, we should determine the range of  for which C2
lies in the region bounded by C1 . Since both 1  sin  , 1  cos are non-negative and periodic
with period 2 , it is equivalent to finding the range such that 1  cos  1  sin  in [0, ) . We
3
7
may further restrict our consideration in [0, 2 ] . The required range is thus (0, )  ( , 2 ] .
4
4
Therefore, the arc length required is


0
[

d
(1  cos  )]2  (1  cos  ) 2 d   sin 2   1  2 cos   cos 2  d
0
d




0
0
2  2 cos  d

4sin 2 d
2


  2sin d
0
2
 4 cos


20
4
Page 9 of 11
Theorem 3.4.
Let r  r ( ) be the polar equation of a curve, a  b be real numbers, then the area S of the
surface generated by revolving the curve about the initial line is given by
2
S   2 r sin 
1
2
 dr 
2

  r d
d



Example 3.6.
Find the surface area of the sphere generated by revolving the curve r  2a cos  , a  0 about the
initial line.
Solution.
The sphere is generated by revolving the part of the curve with domain [0,  ] about the initial line.
Therefore,
S
 2
0
2 (2a cos  ) sin 
 4 a 
 2
0
 4 a 2 
d
(2a cos  )]2  (2a cos  ) 2 d
d
cos  sin  4a 2 sin 2   4a 2 cos 2  d
 2
0
sin 2 d
 2 a 2 cos 2
 4 a
[
 2
0
2
Exercise
1.
Let C1 : r  6cos  , C2 : r  2  2cos  be curves in polar coordinate.
a.
Sketch the two curves.
b.
Find all the intersections of the two curves.
c.
Find the slopes of the curves C1 at the intersection points.
d.
Find the area of the common region bounded by both curves.
Page 10 of 11
4. Solutions to Selected Exercise
Graphs in Polar Coordinates
1.
5
3
4
2
3
1
2
-5
-4
-3
-2
-1
1
1
-1
-2
-1
1
2
3
4
5
-2
-1
-3
-2
figure 11: r  3  3sin( 
figure 10: r  3  2 cos 

)
4
Circle: a  0 or b  0
Limacon: a  0, b  0, a  b
2.
Cardioid: a  0, b  0, a  b
Measurement of Curves in Polar Equation
1.a.
3
2
1
1
2
3
4
5
6
-1
-2
-3
figure 12: C1 and C2 in exercise 4.1

5
b. (3, ), (3, ), (0, 0)
3
3
c.
d.
1

1
5
at (3, ), 
at (3, ), undefined at (0, 0)
3
3
3
3
5
2
Page 11 of 11